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Why does a meromorphic function with limit at infinity continue to have limit at infinity when the poles are removed?
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I wanna prove the following:
If I have a meromorphic function $f$ with finitely many poles $p_1
, ..., p_n$, and it has a limit at infinity (potentially being infinity, as in, the function is meromorphic at infinity), then the entire function $g(z) = f(z)(z-p_1
)^m_1...(z-p_n
)^m_n$ (where the $m_n$ are the order of the poles) also is meromorphic at infinity.
I'm using this to prove holomorphic functions from the sphere to itself are rational, but all the proofs I've found gloss over this part as obvious. If the limit at infinity is infinity or non zero, then clearly $g$ has a pole at infinity, but if the limit is zero then I'm not sure what happens.
complex-analysis meromorphic-functions
edited Jul 24 at 16:37
José Carlos Santos
113k1697176
asked Jul 24 at 16:30
violeta
327110
add a comment |Â
up vote
1
down vote
favorite
I wanna prove the following:
If I have a meromorphic function $f$ with finitely many poles $p_1
, ..., p_n$, and it has a limit at infinity (potentially being infinity, as in, the function is meromorphic at infinity), then the entire function $g(z) = f(z)(z-p_1
)^m_1...(z-p_n
)^m_n$ (where the $m_n$ are the order of the poles) also is meromorphic at infinity.
I'm using this to prove holomorphic functions from the sphere to itself are rational, but all the proofs I've found gloss over this part as obvious. If the limit at infinity is infinity or non zero, then clearly $g$ has a pole at infinity, but if the limit is zero then I'm not sure what happens.
complex-analysis meromorphic-functions
edited Jul 24 at 16:37
José Carlos Santos
113k1697176
asked Jul 24 at 16:30
violeta
327110
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I wanna prove the following:
If I have a meromorphic function $f$ with finitely many poles $p_1
, ..., p_n$, and it has a limit at infinity (potentially being infinity, as in, the function is meromorphic at infinity), then the entire function $g(z) = f(z)(z-p_1
)^m_1...(z-p_n
)^m_n$ (where the $m_n$ are the order of the poles) also is meromorphic at infinity.
I'm using this to prove holomorphic functions from the sphere to itself are rational, but all the proofs I've found gloss over this part as obvious. If the limit at infinity is infinity or non zero, then clearly $g$ has a pole at infinity, but if the limit is zero then I'm not sure what happens.
complex-analysis meromorphic-functions
edited Jul 24 at 16:37
José Carlos Santos
113k1697176
asked Jul 24 at 16:30
violeta
327110
I wanna prove the following:
If I have a meromorphic function $f$ with finitely many poles $p_1
, ..., p_n$, and it has a limit at infinity (potentially being infinity, as in, the function is meromorphic at infinity), then the entire function $g(z) = f(z)(z-p_1
)^m_1...(z-p_n
)^m_n$ (where the $m_n$ are the order of the poles) also is meromorphic at infinity.
I'm using this to prove holomorphic functions from the sphere to itself are rational, but all the proofs I've found gloss over this part as obvious. If the limit at infinity is infinity or non zero, then clearly $g$ has a pole at infinity, but if the limit is zero then I'm not sure what happens.
complex-analysis meromorphic-functions
edited Jul 24 at 16:37
José Carlos Santos
113k1697176
asked Jul 24 at 16:30
violeta
327110
edited Jul 24 at 16:37
José Carlos Santos
113k1697176
edited Jul 24 at 16:37
José Carlos Santos
113k1697176
edited Jul 24 at 16:37
José Carlos Santos
113k1697176
113k1697176
asked Jul 24 at 16:30
violeta
327110
asked Jul 24 at 16:30
violeta
327110
asked Jul 24 at 16:30
violeta
327110
327110
add a comment |Â
add a comment |Â
2 Answers
2
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oldest
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1
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The status of $g$ at infinity is the status of $zmapsto g(frac 1z)$ at $z=0$.
Now $$gleft(frac 1zright) = fleft(frac 1zright)left(frac 1z-p_1
right)^m_1...left(frac 1z-p_n
right)^m_n$$
Which, around zero, is the product of $n+1$ meromorphic functions.
answered Jul 24 at 16:36
Arnaud Mortier
18.9k22159
oh, right, and product of meromorphic being meromorphic is easily proved using the Laurent expansion. thanks :)
– violeta
Jul 24 at 22:01
add a comment |Â
up vote
0
down vote
If two functions are meromorphic at infinity, then so is their product. So, since $f$ and $zmapsto(z-p_1)^m_1ldots(z-p_n)^m_n$ are meromorphic, so is their product.
answered Jul 24 at 16:36
José Carlos Santos
113k1697176
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The status of $g$ at infinity is the status of $zmapsto g(frac 1z)$ at $z=0$.
Now $$gleft(frac 1zright) = fleft(frac 1zright)left(frac 1z-p_1
right)^m_1...left(frac 1z-p_n
right)^m_n$$
Which, around zero, is the product of $n+1$ meromorphic functions.
answered Jul 24 at 16:36
Arnaud Mortier
18.9k22159
oh, right, and product of meromorphic being meromorphic is easily proved using the Laurent expansion. thanks :)
– violeta
Jul 24 at 22:01
add a comment |Â
up vote
1
down vote
accepted
The status of $g$ at infinity is the status of $zmapsto g(frac 1z)$ at $z=0$.
Now $$gleft(frac 1zright) = fleft(frac 1zright)left(frac 1z-p_1
right)^m_1...left(frac 1z-p_n
right)^m_n$$
Which, around zero, is the product of $n+1$ meromorphic functions.
answered Jul 24 at 16:36
Arnaud Mortier
18.9k22159
oh, right, and product of meromorphic being meromorphic is easily proved using the Laurent expansion. thanks :)
– violeta
Jul 24 at 22:01
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The status of $g$ at infinity is the status of $zmapsto g(frac 1z)$ at $z=0$.
Now $$gleft(frac 1zright) = fleft(frac 1zright)left(frac 1z-p_1
right)^m_1...left(frac 1z-p_n
right)^m_n$$
Which, around zero, is the product of $n+1$ meromorphic functions.
answered Jul 24 at 16:36
Arnaud Mortier
18.9k22159
The status of $g$ at infinity is the status of $zmapsto g(frac 1z)$ at $z=0$.
Now $$gleft(frac 1zright) = fleft(frac 1zright)left(frac 1z-p_1
right)^m_1...left(frac 1z-p_n
right)^m_n$$
Which, around zero, is the product of $n+1$ meromorphic functions.
answered Jul 24 at 16:36
Arnaud Mortier
18.9k22159
answered Jul 24 at 16:36
Arnaud Mortier
18.9k22159
answered Jul 24 at 16:36
Arnaud Mortier
18.9k22159
answered Jul 24 at 16:36
Arnaud Mortier
18.9k22159
18.9k22159
oh, right, and product of meromorphic being meromorphic is easily proved using the Laurent expansion. thanks :)
– violeta
Jul 24 at 22:01
add a comment |Â
oh, right, and product of meromorphic being meromorphic is easily proved using the Laurent expansion. thanks :)
– violeta
Jul 24 at 22:01
oh, right, and product of meromorphic being meromorphic is easily proved using the Laurent expansion. thanks :)
– violeta
Jul 24 at 22:01
oh, right, and product of meromorphic being meromorphic is easily proved using the Laurent expansion. thanks :)
– violeta
Jul 24 at 22:01
add a comment |Â
up vote
0
down vote
If two functions are meromorphic at infinity, then so is their product. So, since $f$ and $zmapsto(z-p_1)^m_1ldots(z-p_n)^m_n$ are meromorphic, so is their product.
answered Jul 24 at 16:36
José Carlos Santos
113k1697176
add a comment |Â
up vote
0
down vote
If two functions are meromorphic at infinity, then so is their product. So, since $f$ and $zmapsto(z-p_1)^m_1ldots(z-p_n)^m_n$ are meromorphic, so is their product.
answered Jul 24 at 16:36
José Carlos Santos
113k1697176
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If two functions are meromorphic at infinity, then so is their product. So, since $f$ and $zmapsto(z-p_1)^m_1ldots(z-p_n)^m_n$ are meromorphic, so is their product.
answered Jul 24 at 16:36
José Carlos Santos
113k1697176
If two functions are meromorphic at infinity, then so is their product. So, since $f$ and $zmapsto(z-p_1)^m_1ldots(z-p_n)^m_n$ are meromorphic, so is their product.
answered Jul 24 at 16:36
José Carlos Santos
113k1697176
answered Jul 24 at 16:36
José Carlos Santos
113k1697176
answered Jul 24 at 16:36
José Carlos Santos
113k1697176
answered Jul 24 at 16:36
José Carlos Santos
113k1697176
113k1697176
add a comment |Â
add a comment |Â
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